Question: Solve for $x$ : $ 2|x + 6| + 7 = 4|x + 6| + 4 $
Solution: Subtract $ {2|x + 6|} $ from both sides: $ \begin{eqnarray} 2|x + 6| + 7 &=& 4|x + 6| + 4 \\ \\ {- 2|x + 6|} && {- 2|x + 6|} \\ \\ 7 &=& 2|x + 6| + 4 \end{eqnarray} $ Subtract $4$ from both sides: $ \begin{eqnarray} 7 &=& 2|x + 6| + 4 \\ \\ {- 4} && {- 4} \\ \\ 3 &=& 2|x + 6| \end{eqnarray} $ Divide both sides by ${2}$ $ \dfrac{3} {{2}} = \dfrac{2|x + 6|} {{2}} $ Simplify: $ \dfrac{3}{2} = |x + 6| $ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ -\dfrac{3}{2} = x + 6 $ or $ \dfrac{3}{2} = x + 6 $ Solve for the solution where $x + 6$ is negative: $ - \dfrac{3}{2} = x + 6$ Subtract ${6}$ from both sides: $ \begin{eqnarray} - \dfrac{3}{2} &=& x + 6 \\ \\ {- 6} && {- 6} \\ \\ -\dfrac{3}{2} - 6 &=& x \end{eqnarray} $ Change the ${ - 6}$ to an equivalent fraction with a denominator of $2$ $ - \dfrac{3}{2} {- \dfrac{12}{2}} = x $ $ -\dfrac{15}{2} = x $ Then calculate the solution where $x + 6$ is positive: $ \dfrac{3}{2} = x + 6 $ Subtract ${6}$ from both sides: $ \begin{eqnarray} \dfrac{3}{2} &=& x + 6 \\ \\ {- 6} && {- 6} \\ \\ \dfrac{3}{2} - 6 &=& x \end{eqnarray} $ Change the ${ - 6}$ to an equivalent fraction with a denominator of $2$ $ \dfrac{3}{2} {- \dfrac{12}{2}} = x $ $ -\dfrac{9}{2} = x $ Thus, the correct answer is $x = -\dfrac{15}{2} $ or $x = -\dfrac{9}{2} $.